A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$, and that $f(x) = 1 - |x - 2|$ for $1\leq x \leq 3$. Find the smallest $x$ for which $f(x) = f(2001)$.
Answer: Using the given $f(3x) = 3f(x)$ repeatedly, we have that \[f(2001) = 3f\left(\frac{2001}{3}\right) = 3^2f\left(\frac{2001}{3^2}\right) = \dots = 3^6f\left(\frac{2001}{3^6}\right).\]Since $1 \le 2001/3^6 \le 3,$ we can apply the second part of the definition of $f$ to get \[f(2001) = 3^6\left(1 - \left|\frac{2001}{3^6} - 2\right|\right) = 3 \cdot 3^6 - 2001 = 186.\]Therefore, we want the smallest $x$ for which $f(x) = 186.$ Note that the range of $f(x) $ in the interval $x \in [1, 3]$ is $[0, 1].$ Since $f(3x) = 3f(x)$ for all $x,$ it follows that the range of $f(x)$ in the interval $x \in [3, 9]$ is $[0,3].$ Similarly, for each $k,$ the range of $f(x)$ in the interval $x \in [3^k, 3^{k+1}]$ is $[0, 3^k].$ Therefore, if $f(x) = 186,$ then $3^k \ge 186,$ so $k \ge 5.$

We search the interval $x \in [3^5, 3^6] = [243, 729].$ We want $f(x) = 186,$ and for any $x$ in this interval, we have $f(x) = 3^5f\left(\frac{x}{3^5}\right).$ Therefore, letting $y = \frac{x}{3^5},$ we want $f(y) = \frac{186}{3^5} = \frac{186}{243},$ where $y \in [1, 3].$ That is, \[1 - |y-2| = \frac{186}{243} \implies |y-2| = \frac{57}{243}.\]The smaller of the two solutions to this equation is $y = 2 - \frac{57}{243} = \frac{429}{243}.$ Thus, $x = 3^5y = \boxed{429}.$